Problem: Integrate. $ \int 3\sec(x)\tan(x)\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $\tan x + C$ (Choice B) B $3\sec x + C$ (Choice C) C $3\tan x + C$ (Choice D) D $\sec x + C$
Solution: We need a function whose derivative is $3\sec(x)\tan(x)$. We know that the derivative of $\sec(x)$ is $\sec(x)\tan(x)$, so let's start there: $\dfrac{d}{dx} \sec(x) = \sec(x)\tan(x)$ Now let's multiply by $3$ : $\dfrac{d}{dx} \left[ 3\sec(x) \right]= 3\dfrac{d}{dx} \sec(x) =3\sec(x)\tan(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int 3\sec(x)\tan(x)\,dx =3\sec(x)\, + C$ The answer: $3 \sec(x)\, + C$